Question: A particle moves in the $xy$ -plane so that at any time $t\geq 0$ its coordinates are $x=2t^2-6t$ and $y=-t^3+10t$. What is the magnitude of the particle's velocity vector at $t=2$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $4\sqrt{10}$ (Choice B) B $2\sqrt{2}$ (Choice C) C $7$ (Choice D) D $10$
Answer: Background When solving motion problems, it's important to remember the relationship between the position vector $(x,y)$, the velocity vector $\vec{v}(t)$, and the acceleration vector $\vec{a}(t)$ of the moving particle: $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$ $\vec{a}(t)=\dfrac{d}{dt}\vec{v}(t)=\left(\dfrac{d^2x}{dt^2},\dfrac{d^2y}{dt^2}\right)$ Setting up the math We are given that the particle's coordinates are $x=2t^2-6t$ and $y=-t^3+10t$, which means its position vector is $(2t^2-6t,-t^3+10t)$. We are asked to find the magnitude of the particle's velocity vector at $t=2$. In other words, we need to find $||\vec{v}(2)||$. Finding $\vec{v}(t)$ $\begin{aligned} \vec{v}(t)&=\left(\dfrac{d}{dt}(2t^2-6t),\dfrac{d}{dt}(-t^3+10t)\right) \\\\ &=(4t-6,-3t^2+10) \end{aligned}$ Finding $\vec{v}(2)$ $\begin{aligned} \vec{v}({2})&=(4({2})-6,-3({2})^2+10) \\\\ &=(8-6,-12+10) \\\\ &=(2,-2) \end{aligned}$ Finding $||\vec{v}(2)||$ $\begin{aligned} ||\vec{v}(2)||&=||(C{2},{-2})|| \\\\ &=\sqrt{(C{2})^2+({-2})^2} \\\\ &=2\sqrt{2} \end{aligned}$ In conclusion, the magnitude of the particle's velocity vector at $t=2$ is $2\sqrt{2}$.